Cube \(PQRSTUVW\) has side length \(2\). Point \(M\) is the midpoint of edge \(UT\). Determine the area of \(\triangle MQR\).
We first draw \(RT\).
In \(\triangle RWT\), \(\angle RWT = 90^\circ\) and \(RW = WT = 2\).
By the Pythagorean Theorem in \(\triangle RWT\), \(RT^2= RW^2 + WT^2= 2^2 + 2^2=8\).
Therefore, \(RT = \sqrt{8}\), since \(RT>0\).
\(\triangle MQR\) has base equal to the length of \(QR\), which is \(2\).
Notice that the height of \(\triangle MQR\) is equal to the distance from side \(QR\) of the cube to side \(UT\) of the cube, which is equal to the length of \(RT\) or \(\sqrt{8}\).
Therefore, area of \(\triangle MQR = \dfrac{\text{base}\times \text{height}}{2} = \dfrac{2 \times \sqrt{8}}{2}= \sqrt{8}\) units squared.